two photon annihilation

[phd15@phys.keele.ac.uk]

Question,

this is from Sakurai Advanced QM page 218. For two photon annhiliation
in order to evaluate the differential cross section the reader is
required to evaluate |Mfi|^2. this involves evaluating the trace of a
product of the sort.

Tr[c.a c.b c.d c.e]

where c ~ gamma. Now one particular argument of a trace proportional to
m^2 vanishes.ie

c.k2 c.e1 c.e1 c.k2 = k2^2
= 0

where
k2 ~ kvector of one of the photon 2
e1 ~ polarization vector of photon 1

Can anyone explain why k2^2 = 0 ?

[Cl.Massé]

<phd15@phys.keele.ac.uk> a écrit dans le message de news:
e3t2de$9rn$1@planet.kis.keele.ac.uk

> Question,
>
> this is from Sakurai Advanced QM page 218. For two photon annhiliation
> in order to evaluate the differential cross section the reader is
> required to evaluate |Mfi|^2. this involves evaluating the trace of a
> product of the sort.
>
> Tr[c.a c.b c.d c.e]
>
> where c ~ gamma. Now one particular argument of a trace proportional to
> m^2 vanishes.ie
>
> c.k2 c.e1 c.e1 c.k2 = k2^2
> = 0
>
> where
> k2 ~ kvector of one of the photon 2
> e1 ~ polarization vector of photon 1
>
> Can anyone explain why k2^2 = 0 ?

Because the pseudo-norm of the 4-kvector of a photon is always equal to 0?

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